Question

Proceed as on page $\mathbf{269}$to derive the elementary form of${\mathbf{J}}_{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$given in$\left(27\right)$.

Short answer

The elementary form is ${\mathrm{J}}_{-1/2}\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{x\pi }}}\mathrm{cosx}$.

Step-by-step solution

Define Bessel’s equation.

Let the Bessel equation be${\mathrm{x}}^2\mathrm{y}\text{'}\text{'}+\mathrm{xy}\text{'}+({\mathrm{x}}^2-{\mathrm{n}}^2)\mathrm{y}=0$ . This equation has two linearly independent solutions for a fixed value of . A Bessel equation of the first kind, indicated by ${\mathrm{J}}_{\mathrm{n}}\left(\mathrm{x}\right)$, is one of these solutions that may be derived using Frobinous approach.

$\begin{array}{c}{\mathrm{y}}_1={\mathrm{x}}^{\mathrm{a}}{\mathrm{J}}_{\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)\\ {\mathrm{y}}_2={\mathrm{x}}^{\mathrm{a}}{\mathrm{J}}_{-\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)\end{array}$

At$\mathrm{x}=0$, this solution is regular. The second solution, which is singular at$\mathrm{x}=0$, is represented by${\mathrm{Y}}_{\mathrm{n}}\left(\mathrm{x}\right)$and is calleda Bessel function of the second kind.

${\mathrm{y}}_3={\mathrm{x}}^{\mathrm{a}}\left[\frac{{\mathrm{cosp\pi J}}_{\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)-{\mathrm{J}}_{-\mathrm{p}}\left({\mathrm{bx}}^{\mathrm{c}}\right)}{\mathrm{sinp\pi }}\right]$

Determine the Bessel’s function of first kind.

Let the Bessel’s function of first kind be${\mathrm{J}}_{\mathrm{v}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!\mathrm{\Gamma }(1+\mathrm{v}+\mathrm{n})}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}+\mathrm{v}}$.

Substitute$\mathrm{v}=-1/2$in it yields,

${\mathrm{J}}_{-1/2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!\mathrm{\Gamma }(1-\frac{1}{2}+\mathrm{n})}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}-1/2}$

${\mathrm{J}}_{-1/2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!\mathrm{\Gamma }\left(\frac{1}{2}+\mathrm{n}\right)}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}-1/2}$… (1)

Find the gamma function.

Let the gamma function be,

$\begin{array}{c}\mathrm{n}=0,\mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\mathrm{\pi }}\\ \mathrm{n}=1,\mathrm{\Gamma }\left(\frac{3}{2}\right)=\frac{2!}{2^{2}1!}\sqrt{\mathrm{\pi }}\\ \mathrm{n}=2,\mathrm{\Gamma }\left(\frac{5}{2}\right)=\frac{4!}{2^{4}2!}\sqrt{\mathrm{\pi }}\\ \mathrm{n}=3,\mathrm{\Gamma }\left(\frac{7}{2}\right)=\frac{6!}{2^{6}3!}\sqrt{\mathrm{\pi }}\end{array}$

Then, $\mathrm{\Gamma }\left(\frac{1}{2}+\mathrm{n}\right)=\frac{\left(2\mathrm{n}\right)!}{2^{2\mathrm{n}}\mathrm{n}!}$… (2)

Obtain the elementary form.

Substitute the equation (2) into (1).

$\begin{array}{c}{\mathrm{J}}_{-1/2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{x}!\frac{\left(2\mathrm{n}\right)!}{2^{2\mathrm{n}}{\cancel{}}^{2\mathrm{n}}}\sqrt{\mathrm{\pi }}}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}-1/2}\\ =\frac{1}{\sqrt{\mathrm{\pi }}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{2}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}-1/2}\\ =\sqrt{\mathrm{\pi }}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{2}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\mathrm{x}}^{2\mathrm{n}-1/2}\cdot 2^{-2\mathrm{n}+1/2}\\ =\frac{1}{\sqrt{\mathrm{\pi }}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{2}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\mathrm{x}}^{2\mathrm{n}}\cdot {\mathrm{x}}^{-1/2}\cdot 2^{2\mathrm{\pi }}\cdot 2^{1/2}\\ =\frac{1}{\sqrt{\mathrm{\pi }}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\mathrm{x}}^{2\mathrm{n}}\cdot \sqrt{\frac{2}{\mathrm{x}}}\\ =\sqrt{\frac{2}{\mathrm{x\pi }}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\left(2\mathrm{n}\right)!}{\mathrm{x}}^{2\mathrm{n}}\\ =\sqrt{\frac{2}{\mathrm{x\pi }}}\mathrm{cosx}\end{array}$

Hence verified.